## Lagrangian Pairs

As described in a previous blog, it is possible for to objects to share an orbit if their locations are according to the Lagrange configuration described. (There are other orbit "sharing" configurations, such as horseshoe orbits, not covered herein).

In fact, there are two sets of Lagrangian

Lagrangian moons and planets are able to share an orbit in a stable configuration, located 60° apart. While it is inviting to imagine a system of six worlds equally spaced around a given orbit—and thus all in one another’s L4 and L5 points—studies have shown that without some other

There are a number of possible configurations of planet types as Lagrangian sets, such as an ice giant with two terrestrials at its L4 and L5; a gas giant with two ice giants each leading and trailing it its orbit; a Telluric with two dwarf-planet companions, etc. Sean Raymond at Planetplanet.net [2] has written a truly fascinating series of blog posts in which he explores the various possibilities.

Lagrangian pairs are not double planets (see below), as they do not orbit a barycenter common only to the two of them, but each independently orbits the primary mass component of the stellar system, but both on the same orbit. Also, double planets orbit their common barycenter fairly close to each other—typically within a few tens of their own radii. Lagrangian pairs are much farther apart: indeed, since they are separated by 60° on the same orbit, they are, by definition, as far from one another as each is from the center of the star system. Inhabitants of small enough Lagrangian pair planets might not even be aware of one another until an indigenous culture on one or the other invents telescopy. Asteroid 2010 TK7, which orbits at Earth's L4 point, was only discovered in 2010, and then only by the space-based WISE telescope.

(As it turns out, objects orbiting at Earth's L4 and L5 points would only be visible just before sunrise and/or just after sunset, in much the same way Venus is from Earth. As the diagram shows, those areas of the observing planet in the shaded area would see the companion as a "morning star", rising before the sun, and fading from view as the brightness of the daylight sky drowned it out of view. If the object were

Let's say Earth had an identical twin orbiting at the L5 point.

Using the equation for the small-angle approximation:

Distance between Earth and the twin: 1.496E+08 km

Actual diameter of Earth twin: 12742 km

In fact, there are two sets of Lagrangian

*triples*right here in our own back yard: Saturn's moons Tethys-Telesto-Calypso and Dione-Helene-Polydeuces.*(If you have not previously read my blog on Orbital Libration Points, some of the following may not make a lot of sense.)*

Lagrangian moons and planets are able to share an orbit in a stable configuration, located 60° apart. While it is inviting to imagine a system of six worlds equally spaced around a given orbit—and thus all in one another’s L4 and L5 points—studies have shown that without some other

*strongly*stabilizing influence, such a system would not remain stable, especially for highly elliptical orbits. [1]There are a number of possible configurations of planet types as Lagrangian sets, such as an ice giant with two terrestrials at its L4 and L5; a gas giant with two ice giants each leading and trailing it its orbit; a Telluric with two dwarf-planet companions, etc. Sean Raymond at Planetplanet.net [2] has written a truly fascinating series of blog posts in which he explores the various possibilities.

Lagrangian pairs are not double planets (see below), as they do not orbit a barycenter common only to the two of them, but each independently orbits the primary mass component of the stellar system, but both on the same orbit. Also, double planets orbit their common barycenter fairly close to each other—typically within a few tens of their own radii. Lagrangian pairs are much farther apart: indeed, since they are separated by 60° on the same orbit, they are, by definition, as far from one another as each is from the center of the star system. Inhabitants of small enough Lagrangian pair planets might not even be aware of one another until an indigenous culture on one or the other invents telescopy. Asteroid 2010 TK7, which orbits at Earth's L4 point, was only discovered in 2010, and then only by the space-based WISE telescope.

(As it turns out, objects orbiting at Earth's L4 and L5 points would only be visible just before sunrise and/or just after sunset, in much the same way Venus is from Earth. As the diagram shows, those areas of the observing planet in the shaded area would see the companion as a "morning star", rising before the sun, and fading from view as the brightness of the daylight sky drowned it out of view. If the object were

*very*large and had a high albedo, it might be faintly visible in a clear daylight sky. Also, the object would become visible to about half of the daylight side of the planet during solar eclipses, when the sun's light is blocked ... assuming the planet has at least one moon big enough to completely eclipse the sun.)Let's say Earth had an identical twin orbiting at the L5 point.

Using the equation for the small-angle approximation:

Distance between Earth and the twin: 1.496E+08 km

Actual diameter of Earth twin: 12742 km

... we find that the other planet would have a size in the sky which is ~106.1 times smaller than the size of the full Moon—and that would be for a body the size of Earth!

Considering that the Sun appears about the same size as the full Moon in Earth's sky, and is as far away from Earth as an object would be at Earth's L4 or L5 points, any object at a Lagrange point in Earth's orbit would have to be the size of the Sun in order to appear the size of the full Moon in Earth's sky!

Considering that the Sun appears about the same size as the full Moon in Earth's sky, and is as far away from Earth as an object would be at Earth's L4 or L5 points, any object at a Lagrange point in Earth's orbit would have to be the size of the Sun in order to appear the size of the full Moon in Earth's sky!

Calculating the apparent brightness of our Earth-size planet at L4:

... shows that this twin would have an apparent brightness of ~000009646 watts/m², which is 0.02% the brightness of the Moon, or .0.1523 times the brightness of Venus. It

If such a twin had existed in Earth's orbit, it would surely have been noted by the ancients (who

Once it came under observation using space-based telescopes, it would always be lit as a waxing crescent, though its brightness would likely change due to its axial rotation, differing cloud patterns, and whether land or ocean was visible in the daylit portion (assuming it had oceans).

*would*be 3.5 times brighter than Jupiter and 72 times brighter than Saturn, but it would be invisible in a daytime sky. Were it to be visible, it would be more likely to be noticed than either Jupiter or Saturn, because its position in the*sky*would never change. Were it to be visible at night, its position relative to the background stars*would*change on a nightly basis by 0.986°, almost twice the width of the full Moon.If such a twin had existed in Earth's orbit, it would surely have been noted by the ancients (who

*did*observe the sky during total solar eclipses), and its special nature would have been quickly realized: though it would only appear during an eclipse, its orientation with respect to the Sun in the sky would never change.Once it came under observation using space-based telescopes, it would always be lit as a waxing crescent, though its brightness would likely change due to its axial rotation, differing cloud patterns, and whether land or ocean was visible in the daylit portion (assuming it had oceans).

## Double-Planets

There is currently no official definition of what constitutes a double (or binary) planet. One distinction that would certainly make sense is that both bodies orbit a common barycenter; this at least distinguishes them from co-orbital bodies such as Lagrangian pairs or orbit-exchanging pairs such as Saturn’s moons Epimetheus and Janus.

Some suggested criteria for double-planets include:

It is possible to determine all of these. Let's examine how.

I have described calculating the barycenter of a two-body system in this previous blog, but let's revisit the equation here:

Some suggested criteria for double-planets include:

- Both bodies must qualify individually as planets, per the 2006 IAU definition;
- Both orbit a common center-of-gravity (barycenter), which itself orbits their host star(s);
- The barycenter of their combined orbits
*must lie outside the radius of*;**both**bodies - The ratio of their masses must approach a value of 1.0.

It is possible to determine all of these. Let's examine how.

I have described calculating the barycenter of a two-body system in this previous blog, but let's revisit the equation here:

The benefit to expressing

Using the following values:

Mass of the Earth = 1.0

Mass of the Moon = 0.0123 Earth masses

Average separation between Earth and Moon = 60.336 Earth radii

**in terms of the radius of the primary is that if the value returned for***a**rP ≤ 1.0*, then it is clear that the barycenter lies within the physical radius of the primary.**The Earth-Moon Example**Using the following values:

Mass of the Earth = 1.0

Mass of the Moon = 0.0123 Earth masses

Average separation between Earth and Moon = 60.336 Earth radii

...reveals that the barycenter for the Earth-Moon pair lies within the radius of Earth, and so, technically, this pair cannot be defined as a double-planet. (There is an argument to the contrary; see under the heading "The Tug-Of-War Ratio", below.)

It can happen that the barycenter migrates between lesser-than and greater-than the radius of the primary. Applying the following relation can reveal if this is the case:

**Variation In The Barycenter**It can happen that the barycenter migrates between lesser-than and greater-than the radius of the primary. Applying the following relation can reveal if this is the case:

Applying the Earth-Moon values:

*rP*= 0.73312 Earth radii*Rp*= 1.0 Earth radius*e*= 0.0549The first term is greater than the second but the second term is

*not*greater than the third term, so the relation as a whole does not evaluate as true, so the Earth-Moon barycenter never lies outside the body of the Earth. This argument can also support the contention that the pair do*not*constitute a double planet.## Maximum possible separation

Though I have discussed calculating minimum and maximum

The following will be true:

The maximum possible separation can be calculated by the equation:

*orbital*separations in this previous blog, here I wish to discuss the*for a double-planet pair.***maximum possible separation**The following will be true:

- The more massive of the pair is the primary;
- They will orbit a barycenter which, itself, will orbit the central star(s);
- The barycenter of the double-planet system will lie outside the radius of either body;
- Their orbital eccentricities around the barycenter will be identical;
- They may be closely or widely spaced, limited by the calculation below

The maximum possible separation can be calculated by the equation:

Note that the masses of the two planetsmust beexpressed in units of the mass of the star, which itself is expressed in terms of the Sun's mass.

This is related to the Hill Sphere; if the two bodies are farther apart than the value of

For a double-planet system of two Earth-mass bodies at 1.0 AU from the Sun:

*Om*, then they escape one another’s gravity and orbit the central mass independently (or escape the star system, altogether).For a double-planet system of two Earth-mass bodies at 1.0 AU from the Sun:

...

At a distance of 902.8 million meters, a twin Earth would subtend 0.81˚ in the sky (about 1.6 times the size of the full Moon; it would orbit ~2.35 times farther away than the Moon orbits the Earth, and have an orbital period of ~70.132 days (bearing in mind that the two planets are, in fact, orbiting the barycenter).

*Om*is ~0.006 AU, or ~902.8 million meters; just over 0.6 times Earth’s Hill radius, or ~141.7 times the radius of the Earth. This means that two Earth-mass bodies in orbit around one another would have to remain closer than 902.8 million meters, or the gravity of the Sun would overpower their gravitational attraction on one another. They would then orbit the Sun independently as individual bodies.At a distance of 902.8 million meters, a twin Earth would subtend 0.81˚ in the sky (about 1.6 times the size of the full Moon; it would orbit ~2.35 times farther away than the Moon orbits the Earth, and have an orbital period of ~70.132 days (bearing in mind that the two planets are, in fact, orbiting the barycenter).

The inner limit for such a system would, of course, be the Roche Limit, which for two Earth-density objects works out to 2.44 Earth radii. At a distance of 2.44 Earth radii, using the small angle approximation:

Radius of the Earth: 6731 km (12742 km in diameter)

Distance to twin: 6731 ⨉ 2.44 = 15545.24 km

Radius of the Earth: 6731 km (12742 km in diameter)

Distance to twin: 6731 ⨉ 2.44 = 15545.24 km

... we calculate that the other Earth would subtend ~46.964° in the sky, or 92 times bigger in apparent size than the full Moon!

However, these twin Earths would orbit one another in ~3.8 hours (see below for how to calculate this), and they would be at one another’s Roche limits. If they moved

*any*closer to each other, they would start to break up from gravitational tidal stresses. They would also be tidally locked, each always showing the same hemisphere to the other.## Orbital Periods for double-planets

The orbital period of double planets—as that of moons—can be found by using the following equations:

For convenience, the values for Earth are given above.

Bear in mind that all of the units

If you don't wish to go to the effort to convert everything to Earth-based units, the following equation uses absolute units for the elements of the system. The same equation can be used in any situation where two bodies are in orbit around one another:

Bear in mind that all of the units

*must*be in**Earth units**for the equation to give results in**Earth days**. Thus,*MP*must be the actual mass of the planet divided by Earth’s actual mass;*MS*must be the actual mass of the secondary divided by Earth’s actual mass; and*RS*must be the actual radius of the secondary’s orbit divided by the actual radius of the Earth.**Calculating Orbital Period Without Converting To Earth Units**If you don't wish to go to the effort to convert everything to Earth-based units, the following equation uses absolute units for the elements of the system. The same equation can be used in any situation where two bodies are in orbit around one another:

## Orbital Speed for double-planets

Should there be a need to discover the orbital speed of the members of a double-planet system, it is sadly not as straightforward as finding the orbital speed of a planet around a star.

As the website MathIsFun.com states: “Rather strangely, the perimeter of an ellipse is very difficult to calculate!” [3] They list several approximations, but I’ve chosen to go with the first one they list, which has as its primary limitation the requirement that the semi-minor axis of the ellipse be no less than one-third the length of the semi-major axis.

Since a 3:1 ratio between the semi-major and semi-minor axes results in an eccentricity of 0.94281, and that is an

The following is an equation for calculating the approximate perimeter of an ellipse:

As the website MathIsFun.com states: “Rather strangely, the perimeter of an ellipse is very difficult to calculate!” [3] They list several approximations, but I’ve chosen to go with the first one they list, which has as its primary limitation the requirement that the semi-minor axis of the ellipse be no less than one-third the length of the semi-major axis.

Since a 3:1 ratio between the semi-major and semi-minor axes results in an eccentricity of 0.94281, and that is an

*extremely high*eccentricity for an elliptical orbit (the highest eccentricity of any known moon in the Solar System is that of Nereid, at 0.75 [4]), I consider that the inaccuracies of using this method are acceptable. (If for some reason you have a situation where such an insanely high eccentricity exists in your double-planet or planet-moon pairing, please see the mathisfun.com site for more accurate approximations.)The following is an equation for calculating the approximate perimeter of an ellipse:

Notice that I have not specified units for

The next equation is:

*a*or*b*; this is because the units used will be determined by the desired units of the ultimate answer. If you are using days for the orbital period and you use meters for the axes, your answer will be in meters per day. If you want kilometers per second, it will be necessary to convert the orbital period in days into seconds and use kilometers for the axes.The next equation is:

Let's crunch the numbers for the Moon:

*a*= 3.844E+5 kilometers*b*= 3.838E+5 kilometers (calculated from*a*and the eccentricity of its orbit);*P*= 2.361E+6 secondsDividing by the orbital period in seconds:

which is exactly the same as the official value of 1.022 km/sec [5].

## The "tug-of-war" ratio

The result is interpreted as follows: the larger the number, the greater the ratio of the gravitational pull of the primary on the secondary as compared to the pull of the star. The magnitude of

Below is at table listing the tug-of-war values for various planet-moon pairs:

*FT*, then, determines whether or not the pair are a double planet.Below is at table listing the tug-of-war values for various planet-moon pairs:

The Earth-Moon figure is interesting: it is less than 1.0, which means that the Sun actually has a stronger gravitational pull on the Moon than does the Earth, and that means that—by

In Asimov’s words the Moon is, “…neither a true satellite of the Earth nor a captured one, but is a planet in its own right, moving about the Sun in careful step with the Earth.” [7]

Of course, by modern IAU standards, the Moon would have to be called (ahem) a dwarf planet….

Asimov’s argument is further strengthened by the fact that at no point in its journey around the Sun does the Moon ever travel “backward” relative to the Sun; “It is always ‘falling toward’ the Sun. All the other satellites, without exception, ‘fall away’ from the Sun through part of their orbits, caught as they are by the superior pull of their primary planets – but not the Moon.” [8]

In other words, all of the other satellites in the Solar System have loops in their orbit which, when viewed from above the plane of the Solar System, means that for part of their orbits they travel in the opposite direction of their primary’s orbital motion. The Moon’s orbit does not have any such loops, because the speed at which the pair are orbiting the Sun is faster than any "backward" velocity the moon has in its orbit when it is between the Earth and the Sun.

A very helpful explanation of this is given by Helmer Aslaksen at his website

*Asimov's*reckoning—the Moon is not a satellite of the Earth, but a companion of it (even though we showed earlier that they are*not*a double-planet system, based on the location of their barycenter).In Asimov’s words the Moon is, “…neither a true satellite of the Earth nor a captured one, but is a planet in its own right, moving about the Sun in careful step with the Earth.” [7]

Of course, by modern IAU standards, the Moon would have to be called (ahem) a dwarf planet….

Asimov’s argument is further strengthened by the fact that at no point in its journey around the Sun does the Moon ever travel “backward” relative to the Sun; “It is always ‘falling toward’ the Sun. All the other satellites, without exception, ‘fall away’ from the Sun through part of their orbits, caught as they are by the superior pull of their primary planets – but not the Moon.” [8]

In other words, all of the other satellites in the Solar System have loops in their orbit which, when viewed from above the plane of the Solar System, means that for part of their orbits they travel in the opposite direction of their primary’s orbital motion. The Moon’s orbit does not have any such loops, because the speed at which the pair are orbiting the Sun is faster than any "backward" velocity the moon has in its orbit when it is between the Earth and the Sun.

A very helpful explanation of this is given by Helmer Aslaksen at his website

*Heavenly Mathematics & Cultural Astronomy*[9].1. en.wikipedia.org/wiki/Klemperer_rosette

2. planetplanet.net/2014/05/13/building-the-ultimate-solar-system/

3. www.mathsisfun.com/geometry/ellipse-perimeter.html

4. nineplanets.org/data.html

5. en.wikipedia.org/wiki/Moon

6. Asimov, Isaac (1976).

7. en.wikipedia.org/wiki/Double_planet - cite_note-Asimov-5

8. en.wikipedia.org/wiki/Double_planet - cite_note-PoV-7

9. www.math.nus.edu.sg/aslaksen/teaching/convex.html

2. planetplanet.net/2014/05/13/building-the-ultimate-solar-system/

3. www.mathsisfun.com/geometry/ellipse-perimeter.html

4. nineplanets.org/data.html

5. en.wikipedia.org/wiki/Moon

6. Asimov, Isaac (1976).

*Asimov on Astronomy*. Coronet Books. p125–139. ISBN 0-340-20015-4.7. en.wikipedia.org/wiki/Double_planet - cite_note-Asimov-5

8. en.wikipedia.org/wiki/Double_planet - cite_note-PoV-7

9. www.math.nus.edu.sg/aslaksen/teaching/convex.html